//给你一个由 '1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。 
//
// 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。 
//
// 此外，你可以假设该网格的四条边均被水包围。 
//
// 
//
// 示例 1： 
//
// 
//输入：grid = [
//  ["1","1","1","1","0"],
//  ["1","1","0","1","0"],
//  ["1","1","0","0","0"],
//  ["0","0","0","0","0"]
//]
//输出：1
// 
//
// 示例 2： 
//
// 
//输入：grid = [
//  ["1","1","0","0","0"],
//  ["1","1","0","0","0"],
//  ["0","0","1","0","0"],
//  ["0","0","0","1","1"]
//]
//输出：3
// 
//
// 
//
// 提示： 
//
// 
// m == grid.length 
// n == grid[i].length 
// 1 <= m, n <= 300 
// grid[i][j] 的值为 '0' 或 '1' 
// 
//
// Related Topics 深度优先搜索 广度优先搜索 并查集 数组 矩阵 👍 2266 👎 0


package LeetCode.editor.cn;

/**
 * @author ldltd
 * @date 2023-08-23 00:57:05
 * @description 200.岛屿数量
 */
public class NumberOfIslands{
	 public static void main(String[] args) {
	 	 //测试代码
	 	 Solution solution = new NumberOfIslands().new Solution();

	 }
	 
//力扣代码
//leetcode submit region begin(Prohibit modification and deletion)
class Solution {
    public int numIslands(char[][] grid) {
			int res=0;
		for (int i = 0; i < grid.length; i++) {
			for (int j = 0; j < grid[0].length; j++) {
				if(grid[i][j]=='1'){
					dfs(grid,i,j);
					res++;
				}
			}
		}
		return res;
    }
	private void dfs(char [][] grid,int i,int j){
		if(i<0||i>grid.length-1||j<0||j>grid[0].length-1||grid[i][j]!='1') return;
		grid[i][j]='0';
		dfs(grid,i+1,j);
		dfs(grid,i-1,j);
		dfs(grid,i,j+1);
		dfs(grid,i,j-1);
	}
}
//leetcode submit region end(Prohibit modification and deletion)

}
